我对 MIPS 比较陌生，并且一直在寻找示例问题来扩展我的技能。我面临这个问题：
从另一个数组中减去一个数组的元素并替换第一个数组中的值。
给定数组 a 和 b，每个 10 个字长，对于每个值 i，其中 0 <= i < 10，令 c = A[i] - B[i]。然后，如果 c < 0，则设置 A[i] = 0。否则，设置 A[i] = c。
完成代码，以便正确加载、计算和存储数据值。您必须使用循环来遍历加载和存储
这是我的编码尝试，但是我注意到我$12并没有像我希望的那样分别加载和$13的第一个值。任何帮助将不胜感激！ab
.data 0x10010000
.word 23 # a[0]
.word 6
.word 11
.word 7
.word 44
.word 32
.word 9
.word 16
.word 29
.word 13
.data 0x10010040
.word 6 # b[0]
.word 22
.word 9
.word 1
.word 3
.word 15
.word 10
.word 4
.word 30
.word 8
.text
.globl main
main:
lui $16, 0x1001 # $16 contains the address of a[0]
add $17, $16, 0
ori $17, $17, 0x0040 # $17 contains the address of b[0]
li $9, 0 #initiate iterator value
loop:
lw $12, 0($16) #load register with value of a
nop
lw $13, 0($17) #load register with value of b
nop
li $8, 0 #set c to 0
sub $8, $12, $13 # c = A(i) - B(i)
blt $8, 0, negative # if c is negative branch
nop
li $12, 0
add $12, $12, $8 #else A(i) = c
addi $9, $9, 1 #add 1 to iteration value
sw $12, 0($16) #store value back into a
nop
bne $9, 10, update #check that we have not the last array
nop
b exit #if so finish program
nop
negative:
li $12, 0
sw $12, 0($16) #store value back into a
addi $9, $9, 1 #add 1 to iteration value
beq $9, 10, exit
nop
b loop
nop
update:
add $16, $16, 4 #move to the next value in a
add $17, $17, 4 #move to next value in b
b loop
nop
exit:
b exit
nop